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1 year ago

For the rotation - 660°, find the coterminal angle from 0° < 0 < 360°, thequadrant and the reference angle.

Mathematics

1 answer:

guajiro [1.7K]1 year ago

3 0

Given the angle:

-660°

Let's find the coterminal angle from 0≤θ≤360.

To find the coterminal angle, in the interval given, let's keep adding 360 degrees to the angle until we get the angle in the interval,

We have:

Coterminal angle = -660 + 360 = -300 + 360 = 60°

Therefore, the coterminal angle is 60°.

Since 60 degrees is between 0 to 90 degrees, is is quadrant I.

60 degrees lie in Quadrant I.

Also since it is in quadrant I, the reference angle is still 60 degrees.

ANSWER:

The coterminal angle is 60°, which lies in quadrant I, with a reference angle of 60°

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Rj, this is the solution to question 19:

• 1st period class sold 60 calendars in total

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We calculate the average per week, this way:

• 1st period class sold 60/4 = 15 calendars per week

• 2nd period class sold 123/4 = 30.75 calendars per week

In consequence the multiplication equation that represents this situation is:

15w + 30.75w = 183, where w is the number of weeks Mrs. Hawkins' classes sold calendars.

This is the solution to question 20:

We already know the average number of calendars that 1st and 2nd period classes sold per week. Now, let's calculate the average for 3nd 4th peiod classes, this way:

• 3rd period class sold 89 calendars in total

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Therefore, the average is:

• 3rd period class sold 89/4 = 22.25 calendars per week

• 4th period class sold 126/4 = 31.5 calendars per week

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15 + 30.75 + 22.25 + 31.5 = 99.5 calendars was the average number sold by all her classes in a week.

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1 year ago

Find the least common multiple (LCM) of 8y^6+ 144y^5+ 640y^4 and 2y^4 + 40y^3 + 200y^2.

Mice21 [21]

Answer:

LCM = $8y^{4}(y+ 10)^{2}(y + 8)$

Step-by-step explanation:

Making factors of $8y^{6}+ 144y^{5}+ 640y^{4}$

Taking $8y^{4}$ common:

$\Rightarrow 8y^{4} (y^{2}+ 18y+ 80)$

Using factorization method:

$\Rightarrow 8y^{4} (y^{2}+ 10y + 8y + 80)\\\Rightarrow 8y^{4} (y (y+ 10) + 8(y + 10))\\\Rightarrow 8y^{4} (y+ 10)(y + 8))\\\Rightarrow \underline{2y^{2}} \times 4y^{2} \underline{(y+ 10)}(y + 8)) ..... (1)$

Now, Making factors of $2y^{4} + 40y^{3} + 200y^{2}$

Taking $2y^{2}$ common:

$\Rightarrow 2y^{2} (y^{2}+ 20y+ 100)$

Using factorization method:

$\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10) ............ (2)$

The underlined parts show the Highest Common Factor(HCF).

i.e. HCF is $2y^{2} (y+ 10)$ .

We know the relation between LCM, HCF of the two numbers 'p' , 'q' and the numbers themselves as:

$HCF \times LCM = p \times q$

Using equations (1) and (2): $\Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)$

Hence, LCM = $8y^{4}(y+ 10)^{2}(y + 8)$

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3 years ago

If you flip a coin 150 times about how many times would you expect to get heads

meriva

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3 years ago

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Answer:

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